Question

The sum to $35terms$ of the series $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+...$ is $S$. Then value of $\frac{6}{7}S$ is

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

The correct option is B $5$

Given series $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+........$

$nth$ term of the series is $\frac{(2n+1)}{\frac{n(n+1)(2n+1)}{6}}$

$=\frac{6}{n(n+1)}$

$S=6{\sum }_{n=1}^{35}\frac{1}{n(n+1)}=6{\sum }_{n=1}^{35}(\frac{1}{n}-\frac{1}{n+1})=6\left[\right(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4}+......(\frac{1}{35}-\frac{1}{36})\left)\right]=6[1-\frac{1}{36}]=6\times \frac{35}{36}=\frac{35}{6}\frac{6}{7}S=\frac{6}{7}\times \frac{35}{6}=5$